Question

A soap bubble of radius $R$ is surrourded by another soap bubble of radius $2R$, as shown. Take surface tension$=S$. Then, the pressure inside the smaller soap bubble, in excess of the atmosphere pressure will be:

• A. $\frac{4S}{R}$
• B. $\frac{6S}{R}$
• C. $\frac{8S}{R}$
• D. $\frac{3S}{R}$

Answer 1

The correct option is B $\frac{6S}{R}$

According to Laplace equation, excess pressure created with surface tension of the spherical surface of the liquid is equal to $△p=\frac{2S}{R}.$

In case of soap bubbles the excess pressure of air inside them is doubled due to the pressure of two interface are inside and the outside ${△}_{pB}=\frac{4S}{R}.$

Excess pressure of air inside the bigger bubble ${△}_{pS}=\frac{4S}{2R}=\frac{2S}{R}$

Excess pressure of the air inside the smaller bubble will be ${△}_{pS}=\frac{4S}{R}$

Air pressure difference between the smaller bubble and the atmosphere will be equal to the sum of excess pressure inside the bigger and smaller bubble ${△}_p={△}_{pS}+{△}_{pS}$

$\Rightarrow \frac{2S}{R}+\frac{4S}{R}=\frac{6S}{R}$

Hence, the answer is $\frac{6S}{R}.$